The missing particle in the following nuclear equation is to be determined. M 12 22 g → N 11 22 a + _ Concept Introduction: When atomic number of element decreases by 1, the type of emission is called positron emission. During positron emission, a proton changes into a neutron and emits a positron in the process. Thus, positron emission is a consequence of emission of positron from an unstable nucleus. The mass of positron is same as that of electron, the charge of positron is 1 + . The symbol positron is denoted as: e + 1 0 .

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Chapter 17, Problem 3SAQ

Interpretation Introduction

Interpretation: The missing particle in the following nuclear equation is to be determined.

M1222g→N1122a+_

Concept Introduction: When atomic number of element decreases by 1, the type of emission is called positron emission. During positron emission, a proton changes into a neutron and emits a positron in the process. Thus, positron emission is a consequence of emission of positron from an unstable nucleus. The mass of positron is same as that of electron, the charge of positron is 1+. The symbol positron is denoted as:e+10.

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Chapter 17, Problem 3SAQ

Interpretation Introduction

Interpretation: The missing particle in the following nuclear equation is to be determined.

M1222g→N1122a+_

Concept Introduction: When atomic number of element decreases by 1, the type of emission is called positron emission. During positron emission, a proton changes into a neutron and emits a positron in the process. Thus, positron emission is a consequence of emission of positron from an unstable nucleus. The mass of positron is same as that of electron, the charge of positron is 1+. The symbol positron is denoted as:e+10.

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Answer 3

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Chapter 17, Problem 3SAQ

Interpretation Introduction

Interpretation: The missing particle in the following nuclear equation is to be determined.

M1222g→N1122a+_

Concept Introduction: When atomic number of element decreases by 1, the type of emission is called positron emission. During positron emission, a proton changes into a neutron and emits a positron in the process. Thus, positron emission is a consequence of emission of positron from an unstable nucleus. The mass of positron is same as that of electron, the charge of positron is 1+. The symbol positron is denoted as:e+10.

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Answer 4

Question

Chapter 17, Problem 3SAQ

Interpretation Introduction

Interpretation: The missing particle in the following nuclear equation is to be determined.

M1222g→N1122a+_

Concept Introduction: When atomic number of element decreases by 1, the type of emission is called positron emission. During positron emission, a proton changes into a neutron and emits a positron in the process. Thus, positron emission is a consequence of emission of positron from an unstable nucleus. The mass of positron is same as that of electron, the charge of positron is 1+. The symbol positron is denoted as:e+10.

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Answer 5

Chapter 17, Problem 3SAQ

Interpretation Introduction

Interpretation: The missing particle in the following nuclear equation is to be determined.

M1222g→N1122a+_

Concept Introduction: When atomic number of element decreases by 1, the type of emission is called positron emission. During positron emission, a proton changes into a neutron and emits a positron in the process. Thus, positron emission is a consequence of emission of positron from an unstable nucleus. The mass of positron is same as that of electron, the charge of positron is 1+. The symbol positron is denoted as:e+10.

Answer 6

Chapter 17, Problem 3SAQ

Interpretation Introduction

Interpretation: The missing particle in the following nuclear equation is to be determined.

M1222g→N1122a+_

Concept Introduction: When atomic number of element decreases by 1, the type of emission is called positron emission. During positron emission, a proton changes into a neutron and emits a positron in the process. Thus, positron emission is a consequence of emission of positron from an unstable nucleus. The mass of positron is same as that of electron, the charge of positron is 1+. The symbol positron is denoted as:e+10.

Answer 7

Chapter 17, Problem 3SAQ

Interpretation Introduction

Interpretation: The missing particle in the following nuclear equation is to be determined.

M1222g→N1122a+_

Concept Introduction: When atomic number of element decreases by 1, the type of emission is called positron emission. During positron emission, a proton changes into a neutron and emits a positron in the process. Thus, positron emission is a consequence of emission of positron from an unstable nucleus. The mass of positron is same as that of electron, the charge of positron is 1+. The symbol positron is denoted as:e+10.

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Ch. 17 - Prob. 1SAQ Ch. 17 - Prob. 2SAQ Ch. 17 - Prob. 3SAQ Ch. 17 - Prob. 4SAQ Ch. 17 - Prob. 5SAQ Ch. 17 - Prob. 6SAQ Ch. 17 - Prob. 7SAQ Ch. 17 - Prob. 8SAQ Ch. 17 - Prob. 9SAQ Ch. 17 - Prob. 10SAQ

Solution Summary: The author explains that the missing particle in the nuclear equation is _+10e.

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